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Matter & Change Chapter 2. A. Describing Matter Understanding matter begins with observation Matter is anything that has mass and takes up space Chemistry.

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Presentation on theme: "Matter & Change Chapter 2. A. Describing Matter Understanding matter begins with observation Matter is anything that has mass and takes up space Chemistry."— Presentation transcript:

1 Matter & Change Chapter 2

2 A. Describing Matter Understanding matter begins with observation Matter is anything that has mass and takes up space Chemistry – the study of matter and the changes it undergoes

3 B. Four States of Matter Solids particles vibrate but cant move around fixed shape fixed volume Virtually incompressible

4 B. Four States of Matter Liquids particles can move around but are still close together variable shape fixed volume Virtually incompressible

5 B. Four States of Matter Gases particles can separate and move throughout container variable shape variable volume Easily compressed Vapor = gaseous state of a substance that is a liquid or solid at room temperature

6 B. Four States of Matter Plasma particles collide with enough energy to break into charged particles (+/-) gas-like, variable shape & volume stars, fluorescent light bulbs, TV tubes

7 II. Properties & Changes in Matter Extensive vs. Intensive Physical vs. Chemical

8 A. Physical Properties Physical Property can be observed without changing the identity of the substance

9 B. Physical Properties Physical properties can be described as one of 2 types: Extensive Property depends on the amount of matter present (example: length) Intensive Property depends on the identity of substance, not the amount (example: scent)

10 C. Extensive vs. Intensive Examples: boiling point volume mass density conductivity intensive extensive intensive

11 D. Physical Changes Physical Change changes the form of a substance without changing its identity properties remain the same Examples: cutting a sheet of paper, breaking a crystal, all phase changes

12 D. Phase Changes – Physical Evaporation = Condensation = Melting = Freezing = Sublimation = Liquid -> Gas Gas -> Liquid Solid -> Liquid Liquid -> Solid Solid -> Gas

13 E. Chemical Properties Chemical Property describes the ability of a substance to undergo changes in identity

14 F. Physical vs. Chemical Properties Examples: melting point flammable density magnetic tarnishes in air physical chemical physical chemical

15 G. Chemical Changes Process that involves one or more substances changing into a new substance Commonly referred to as a chemical reaction New substances have different compositions and properties from original substances

16 G. Chemical Changes Signs of a Chemical Change change in color or odor formation of a gas formation of a precipitate (solid) change in light or heat

17 H. Physical vs. Chemical Changes Examples: rusting iron dissolving in water burning a log melting ice grinding spices chemical physical chemical physical

18 What Type of Change?

19 What Type of Change?

20 I. Law of Conservation of Mass Although chemical changes occur, mass is neither created nor destroyed in a chemical reaction Mass of reactants equals mass of products mass reactants = mass products A + B C

21 I. Conservation of Mass In an experiment, 10.00 g of red mercury (II) oxide powder is placed in an open flask and heated until it is converted to liquid mercury and oxygen gas. The liquid mercury has a mass of 9.26 g. What is the mass of the oxygen formed in the reaction? Mercury (II) oxide mercury + oxygen Mmercury(II) oxide = 10.00 g Mmercury = 9.26 Moxygen = ? GIVEN: Mercury (II) oxide mercury + oxygen M mercury(II) oxide = 10.00 g M mercury = 9.86 g M oxygen = ? WORK : 10.00 g = 9.86 g + m oxygen M oxygen = (10.00 g – 9.86 g) M oxygen = 0.74 g mass reactants = mass products

22 III. Classification of Matter (pp. 80-87) Matter Flowchart Pure Substances Mixtures

23 A. Matter Flowchart MATTER Can it be physically separated? Homogeneous Mixture (solution) Heterogeneous MixtureCompoundElement MIXTUREPURE SUBSTANCE yes no Can it be chemically decomposed? noyes Is the composition uniform? noyes

24 A. Matter Flowchart Examples: graphite pepper sugar (sucrose) paint soda element hetero. mixture compound hetero. mixture solution

25 B. Pure Substances Element composed of identical atoms EX: copper wire, aluminum foil

26 B. Pure Substances Compound composed of 2 or more elements in a fixed ratio properties differ from those of individual elements EX: table salt (NaCl)

27 C. Mixtures Variable combination of 2 or more pure substances. HeterogeneousHomogeneous

28 C. Mixtures Solution homogeneous very small particles particles dont settle EX: rubbing alcohol

29 C. Mixtures Heterogeneous medium-sized to large-sized particles particles may or may not settle EX: milk, fresh- squeezed lemonade

30 C. Mixtures Examples: tea muddy water fog saltwater Italian salad dressing Answers: Solution Heterogeneous Solution Heterogeneous

31 III. Density Fun Formula Simple Calculations Dimensional Analysis

32 A. Derived Units Combination of base units Volume – length length length 1 cm 3 = 1 mL 1 dm 3 = 1 L Density – mass per unit volume (g/cm 3 ) D = MVMV D M V Broken Heart

33 B. Density Calculations An object has a volume of 825 cm 3 and a density of 13.6 g/cm 3. Find its mass. GIVEN: V = 825 cm 3 D = 13.6 g/cm 3 M = ? WORK : M = DV M = (13.6 g/cm 3 )(825cm 3 ) M = 11,220 g M = 11,200 g D M V

34 B. Density Calculations A liquid has a density of 0.87 g/mL. What volume is occupied by 25 g of the liquid? GIVEN: D = 0.87 g/mL V = ? M = 25 g D M V WORK : V = M D V = 25 g 0.87 g/mL V = 28.7 mL = 29 mL

35 B. Density Calculations You have a sample with a mass of 620 g and a volume of 753 cm 3. Find its density. GIVEN: M = 620 g V = 753 cm 3 D = ? D M V WORK : D = M V D = 620 g 753 cm 3 D = 0.82 g/cm 3

36 C. Density Calculations with DA Used when units do not agree Conversions must be made before using formula D = MVMV g cm 3

37 You have 3.10 pounds of gold. Find its volume in cm 3 if the density of gold is 19.3 g/cm 3. lbcm 3 3.10 lb 1 kg 2.2 lb = 73.0 cm 3 1000 g 1 kg 1 cm 3 19.3 g C. Density Calculations with DA

38 You have 0.500 L of water. Find its mass in ounces if the density of water is 1.00 g/cm 3. Loz 0.500 L 1000 mL 1L = 17.6 oz 1.00g 1 cm 3 C. Density Calculations with DA 1 cm 3 1 mL 1 kg 1000 g 16 oz 1lb 2.2 lbs 1kg

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